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abc=a+b+c => 1 = 1/ab + 1/bc + 1/ac
2 = 1/a+1/b+1/c => 4 = 1/a^2 + 1/b^2 + 1/c^2 + 2/ab + 2/ac + 2/cb
=> 4 = 1/a^2 + 1/b^2 + 1/c^2 + 2(1/ab + 1/ac + 1/bc) = M + 2
=> M = 4 - 2 = 2
Mk làm bài đầu thôi,sáng nay mk làm cái tt cho
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\)\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c}{abc}+\frac{a}{abc}+\frac{b}{abc}\right)=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\frac{a+b+c}{abc}=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\) (do a+b+c = abc)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(a^3+b^3+c^3=3abc\)
<=> \(a^3+b^3+c^3-3abc=0\)
<=> \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
<=> \(\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
đến đây ez tự làm nốt nhé, ko ra ib mk
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ab+ac}{abc}=2\)
\(\frac{bc+ab+ac}{a+b+c}=2\Leftrightarrow bc+ab+ac=2\left(a+b+c\right)\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}\)( * )
Để \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)thì \(2\left(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}\right)=2\Leftrightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=1\)
\(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{a^2bc+bac^2+ab^2c}{\left(abc\right)^2}=\frac{abc\left(a+b+c\right)}{\left(abc\right)^2}=\frac{a+b+c}{abc}\)
mà a + b + c = abc \(\Rightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{abc}{abc}=1\Leftrightarrow\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\)
thay \(\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\) vào ( * ) ta được \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\left(đpcm\right)\)
\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{bc.ac+ab.ac+ab.bc}{ab.bc.ac}\)
\(=\frac{abc.\left(a+b+c\right)}{a^2b^2c^2}=\frac{a+b+c}{abc}=1\left(\text{vì }a+b+c=abc\right)\)
\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\text{ từ}\left(1\right)\)
Vậy ...
Tham khảo: Câu hỏi của Nguyễn Thị Nhàn - Toán lớp 8 - Học toán với OnlineMath
Học tốt=)
tth : mẫu nó khác bạn nhé
- mẫu nó là 2bc 2ac 2ab
mẫu mk ko có nhân 2
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c